CBSE Class XI

Question 13:

In a survey of  600 students in a school,  150 students were found to be taking tea and 225 taking coffee,  100 were taking both tea and coffee. Find how many students were taking neither tea nor coffee?

Answer

Let U be the set of all students who took part in the survey. Let T be the set of students taking tea.

Let C be the set of students taking coffee.

Accordingly, n(U) = 600, n(T) = 150, n(C) = 225, n(T ∩ C) = 100

To find: Number of student taking neither tea nor coffee i.e., we have to find n(T’∩ C’). n(T’ ∩ C’) = n(T ∪ C)’

= n(U) – n(T ∪ C)

= n(U) – [n(T) + n(C) – n(T ∩ C)]

= 600 – [150 + 225 –100]

= 600 – 275

= 325

Hence, 325 students were taking neither tea nor coffee.

Question 12:

Find sets A, B and C such that A ∩ B, B ∩ C and A ∩ C are non-empty sets and A ∩ B ∩ C = Φ.

Answer

Let A = {0, 1}, B = {1, 2}, and C = {2, 0}.

Accordingly, A ∩ B = {1}, B ∩ C = {2}, and A ∩ C = {0}. ∴ A ∩ B, B ∩ C, and A ∩ C are non-empty.

However, A ∩ B ∩ C = Φ

Question 11:

Let A and B be sets. If A ∩ X = B ∩ X = Φ and A ∪ X = B ∪ X for some set X, show that A = B.

(Hints A = A ∩ (A ∪ X), B = B ∩ (B ∪ X) and use distributive law)

Answer

Let A and B be two sets such that A ∩ X = B ∩ X = f and A ∪ X = B ∪ X for some set X. To show: A = B

It can be seen that

A = A ∩ (A ∪ X) = A ∩ (B ∪ X) [A ∪ X = B ∪ X] = (A ∩ B) ∪ (A ∩ X) [Distributive law]

= (A ∩ B) ∪ Φ [A ∩ X = Φ]

= A ∩ B … (1)

Now, B = B ∩ (B ∪ X)

= B ∩ (A ∪ X) [A ∪ X = B ∪ X]

= (B ∩ A) ∪ (B ∩ X) [Distributive law] 

= (B ∩ A) ∪ Φ [B ∩ X = Φ]

= B ∩ A

= A ∩ B … (2)

Hence, from (1) and (2), we obtain A = B.

Question 10:

Show that A ∩ B = A ∩ C need not imply B = C.

Answer

Let A = {0, 1}, B = {0, 2, 3}, and C = {0, 4, 5}

Accordingly, A ∩ B = {0} and A ∩ C = {0}

Here, A ∩ B = A ∩ C = {0}

However, B ≠ C [2 ∈ B and 2 ∉ C]

Question 9:

Using properties of sets show that 

(i) A ∪ (A ∩ B) = A

(ii) A ∩ (A ∪ B) = A.

Answer

(i) To show: A ∪ (A ∩ B) = A

We know that

A ⊂ A

A ∩ B ⊂ A

 ∴ A ∪ (A ∩ B) ⊂ A … (1)

Also,   A ⊂ A ∪ (A ∩ B) … (2)

∴ From (1) and (2), A ∪ (A ∩ B) = A

(ii) To show: A ∩ (A ∪ B) = A

A ∩ (A ∪ B) = (A ∩ A) ∪ (A ∩ B)

= A ∪ (A ∩ B)

= A {from (1)}

Question 8:

Show that for any sets A and B,

A = (A ∩ B) ∪ (A − B) and A ∪ (B − A) = (A∪ B)

Answer

To show: A = (A ∩ B) ∪ (A − B)

Let x ∈ A

We have to show that x ∈ (A ∩ B) ∪ (A − B)

Case I

x ∈ A ∩ B

Then, x ∈ (A ∩ B) ⊂ (A ∪ B) ∪ (A − B)

Case II

x ∉ A ∩ B

⇒ x ∉ A or x ∉ B

∴ x ∉ B [x ∉ A]

∴ x ∉ A − B ⊂ (A ∪ B) ∪ (A − B)

∴ A ⊂ (A ∩ B) ∪ (A − B) … (1)

It is clear that

A ∩ B ⊂ A and (A − B) ⊂ A

∴ (A ∩ B) ∪ (A − B) ⊂ A … (2)

From (1) and (2), we obtain 

A = (A ∩ B) ∪ (A − B)

To prove: A ∪ (B − A) ⊂ A ∪ B

Let x ∈ A ∪ (B − A)

⇒ x ∈ A or x ∈ (B − A)

⇒ x ∈ A or (x ∈ B and x ∉ A)

⇒ (x ∈ A or x ∈ B) and (x ∈ A or x ∉ A)

⇒ x ∈ (A ∪ B)

∴ A ∪ (B − A) ⊂ (A ∪ B) … (3)

Next, we show that (A ∪ B) ⊂ A ∪ (B − A).

Let y ∈ A ∪ B

⇒ y ∈ A or y ∈ B

⇒ (y ∈ A or y ∈ B) and (y ∈ A or y ∉ A)

⇒ y ∈ A or (y ∈ B and y ∉ A)

⇒ y ∈ A ∪ (B − A)

∴ A ∪ B ⊂ A ∪ (B − A) … (4)

Hence, from (3) and (4), we obtain A ∪ (B − A) = A ∪B.

Question 7:

Is it true that for any sets A and B, P (A) ∪ P (B) = P (A ∪ B)? Justify your answer.

Answer

False 

Let A = {0, 1} and B = {1, 2}

∴ A ∪ B = {0, 1, 2}

P(A) = {Φ, {0}, {1}, {0, 1}}

P(B) = {Φ, {1}, {2}, {1, 2}}

P(A ∪ B) = {Φ, {0}, {1}, {2}, {0, 1}, {1, 2}, {0, 2}, {0, 1, 2}} P(A) ∪ P(B) = {Φ, {0}, {1}, {0, 1}, {2}, {1, 2}}

∴ P(A) ∪ P(B) ≠ P(A ∪ B)

Question 6:

Assume that P (A) = P (B). Show that A = B.

Answer

Let P(A) = P(B)

To show: A = B

Let x ∈ A

A ∈ P(A) = P(B)

∴ x ∈ C, for some C ∈ P(B)

Now, C ⊂ B

∴ x ∈ B

∴ A ⊂ B

Similarly, B ⊂ A

∴ A = B

Question 5:

Show that if A ⊂ B, then C−B ⊂ C − A.

Answer

Let A ⊂ B

To show: C − B ⊂ C − A

Let x ∈ C − B

⇒ x ∈ C and x ∉ B

⇒ x ∈ C and x ∉ A [A ⊂ B]

⇒ x ∈ C − A

∴ C − B ⊂ C – A

Question 4: 

Show that the following four conditions are equivalent:

(i) A ⊂ B                                (ii) A – B = Φ

(iii) A ∪ B = B                     (iv) A ∩ B = A

Answer

First, we have to show that (i) ⇔ (ii).

Let A ⊂ B

To show: A−B ≠ Φ

If possible, suppose A − B ≠ Φ

This means that there exists x ∈ A, x ≠ B, which is not possible as A ⊂ B. ∴ A −B = Φ

∴ A ⊂ B ⇒ A − B = Φ

Let A− B = Φ

To show: A ⊂ B

Let x ∈ A

Clearly, x ∈ B because if x ∉ B, then A − B ≠ Φ ∴ A − B = Φ ⇒ A ⊂ B

∴ (i) ⇔ (ii)

Let A ⊂ B

To show: A ∪ B= B

Clearly, B ⊂ A ∪ B

Let x ∈ A ∪ B

⇒ x ∈ A or x ∈ B

Case I : x ∈ A

⇒ x ∈ B                     [∵ A ⊂ B]

∴ A ∪ B ⊂ B

Case II: x ∈ B

Then, A ∪ B = B

Conversely, let A ∪ B = B

Let x ∈ A

⇒ x ∈ A ∪ B              [∵ A ⊂ A ∪ B]

⇒ x ∈ B                     [∵ A ∪ B = B]

∴ A ⊂ B

Hence, (i) ⇔ (iii)

Now, we have to show that (i) ⇔ (iv).

Let A ⊂ B

Clearly A ∩ B ⊂ A

Let x ∈ A

We have to show that x ∈ A ∩ B

As A ⊂ B, x ∈ B

∴ x ∈ A ∩ B

∴ A ⊂ A ∩ B

Hence, A = A ∩ B

Conversely, suppose A ∩ B = A

Let x ∈ A

⇒ x ∈ A ∩ B

⇒ x ∈ A and x ∈ B

∴ A ⊂ B

Hence, (i) ⇔ (iv).

Question 3:

Let A, B and C be the sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C. show that B = C.

Answer

Let, A, B and C be the sets such that A ∪ B = A ∪ C and A ∩ B = A ∩ C.

To show: B = C

Let x ∈ B

⇒ x ∈ A ∪ B                        [B⊂ A ∪ B]

⇒ x ∈ A ∪ C                        [A ∪ B= A ∪ C]

⇒ x ∈ A or x ∈ C

Case I

x ∈ A

Also, x ∈ B

⇒ x ∈ A ∩ C [∵ A ∩ B = A ∩ C]

∴ x ∈ A and x ∈ C
∴ x ∈ C

∴ B ⊂ C

Similarly, we can show that C ⊂ B.

∴ B = C

Question 2:

In each of the following, determine whether the statement is true or false.

If it is true, prove it. If it is false, give an example.

(i) If x ∈ A and A ∈ B, then x ∈ B
(ii) If A ⊂ B and B ∈ C, then A ∈ C
(iii) If A ⊂ B and B ⊂ C, then A ⊂ C
(iv) If A ⊄ B and B ⊄ C, then A ⊄ C

(v) If x ∈ A and A ⊄ B, then x ∈ B
(vi) If A ⊂ B and x ∉ B, then x ∉ A

Answer

(i) False

Let A = {1, 2} and B = {1, {1, 2}, {3}}

Now, 2∈{1, 2} and {1, 2} ∈ {{3}, 1, {1, 2}}

∴ A ∈ B

However, 2 ∉ {{3}, 1, {1, 3}}

(ii) False

Let A = {2}, B = {0, 2} and C = {1, {0, 2}, 3}

As A ⊂ B
B ∈ C

However, A ∉ C

(iii) True

Let A ⊂ B and B ⊂ C.
Let x ∈ A

⇒ x ∈ B      [∵ A ⊂ B]

⇒ x ∈ C         [∵ B ⊂ C]

∴ A ⊂ C

(iv) False

Let, A {1, 2}, B = {0, 6, 8}, and C = {0, 1, 2, 6, 9}

Accordingly, A ⊄ B and B ⊄ C.

However, A ⊂ C

(v) False

Let A = {3, 5, 7} and B = {3, 4, 6}
Now, 5 ∈ A and A ⊄ B

However, 5 ∉ B

(vi) True

Let A ⊂ B and x ∉ B.
To show: x ∉ A

If possible, suppose x ∈ A.

Then, x ∈ B, which is a contradiction as x ∉ B ∴x ∉ A

Question 1:

Decide, among the following sets, which sets are subsets of one and another: A = {x: x ∈ R and x satisfy x² – 8x + 12 = 0}, B = {2, 4, 6}, C = {2, 4, 6, 8…}, D = {6}.

Answer

A = {x: x ∈ R and x satisfies x² – 8x + 12 = 0}

2 and 6 are the only solutions of x² – 8x + 12 = 0. ∴ A = {2, 6}

B = {2, 4, 6}, C = {2, 4, 6, 8 …}, D = {6}

∴ D ⊂ A ⊂ B ⊂ C

Hence, A ⊂ B, A ⊂ C, B ⊂ C, D ⊂ A, D ⊂ B, D ⊂ C

Question 8: 

In a committee, 50 people speak French,  20 speak Spanish and  10 speak both Spanish and French. How many speak at least one of these two languages?

Let F be the set of people in the committee who speak French, and S be the set of people in the committee who speak Spanish

∴ n(F) = 50, n(S) = 20, n(S ∩ F) = 10

We know that:

n(S ∪ F) = n(S) + n(F) − n(S ∩ F)

= 20 + 50 − 10

= 70 − 10 = 60

Thus, 60 people in the committee speak at least one of the two languages.

Question 7:

In a group of 65 people, 40 like cricket,  10 like both cricket and tennis. How many like tennis only and not cricket? How many like tennis?

Answer 

Let C denote the set of people who like cricket, and T denote the set of people who like tennis

∴ n(C ∪ T) = 65, n(C) = 40, n(C ∩ T) = 10
We know that:

n(C ∪ T) = n(C) + n(T) − n(C ∩ T)
∴ 65 = 40 + n(T) − 10

⇒ 65 = 30 + n(T)

⇒ n(T) = 65 − 30 = 35

Therefore, 35 people like tennis.
Now,

(T − C) ∪ (T ∩ C) = T
Also,

(T − C) ∩ (T ∩ C) = Φ

∴ n (T) = n (T − C) + n (T ∩ C)
⇒ 35 = n (T − C) + 10

⇒ n (T − C) = 35 − 10 = 25

Thus, 25 people like only tennis.

Question 6:

In a group of 70 people, 37 like coffee, 52 like tea, and each person likes at least one of the two drinks. How many people like both coffee and tea?

Answer

Let C denote the set of people who like coffee, and T denote the set of people who like tea
n(C ∪ T) = 70, n(C) = 37, n(T) = 52
We know that:

n(C ∪ T) = n(C) + n(T) − n(C ∩ T)
∴ 70 = 37 + 52 − n(C ∩ T)
⇒ 70 = 89 − n(C ∩ T)
⇒ n(C ∩ T) = 89 − 70 = 19

Thus, 19 people like both coffee and tea.

Question 5:

If X and Y are two sets such that X has 40 elements, X ∪Y has 60 elements and X ∩Y has 10 elements, how many elements does Y have?

Answer

It is given that:

n(X) = 40, n(X ∪ Y) = 60, n(X ∩ Y) = 10
We know that:

n(X ∪ Y) = n(X) + n(Y) − n(X ∩ Y)
∴ 60 = 40 + n(Y) −10

∴ n(Y) = 60 − (40 − 10) = 30

Thus, the set Y has 30 elements.

Question 4:

If S and T are two sets such that S has 21 elements, T has 32 elements, and S ∩ T has 11 elements, how many elements does S ∪ T have?

Answer

It is given that:

n(S) = 21, n(T) = 32, n(S ∩ T) = 11
We know that:

n (S ∪ T) = n (S) + n (T) − n (S ∩ T)

∴ n (S ∪ T) = 21 + 32 − 11 = 42
Thus, the set (S ∪ T) has 42 elements.

Question 3:

In a group of  400 people,  250 can speak Hindi and  200 can speak English. How many people can speak both Hindi and English?

Answer

Let H be the set of people who speak Hindi, and E be the set of people who speak English

∴ n(H ∪ E) = 400, n(H) = 250, n(E) = 200

n(H ∩ E) = ?

We know that:

n(H ∪ E) = n(H) + n(E) − n(H ∩ E)

∴ 400 = 250 + 200 − n(H ∩ E)

⇒ 400 = 450 − n(H ∩ E)

⇒ n(H ∩ E) = 450 – 400

∴ n(H ∩ E) = 50

Thus, 50 people can speak both Hindi and English.

Question 2:

If X and Y are two sets such that X ∪Y has 18 elements, X has 8 elements and Y has 15 elements; how many elements does X ∩Y have?

Answer

It is given that:

n(X ∩ Y) = ?

We know that:

n(X ∪ Y) = n(X) + n(Y) – n(X ∩ Y)

∴ 18 = 8 + 15 – n(X ∩ Y)

⇒ n(X ∩ Y) = 23 – 18 = 5

∴ n(X ∩ Y) = 5